Ideal Gas Equation - Physics

5.2 Pressure of a Gas Gases exert pressure on any surface with which they come in contact, because gas molecules are constantly in motion. We humans have adapted so well physiologically to the pressure of the air around us that we are usually unaware of it, perhaps as fish are not conscious of the w...

5.2 Pressure of a Gas Gases exert pressure on any surface with which they come in contact, because gas molecules are constantly in motion. We humans have adapted so well physiologically to the pressure of the air around us that we are usually unaware of it, perhaps as fish are not conscious of the water’s pressure on them. It is easy to demonstrate atmospheric pressure. One everyday example is the ability to drink a liquid through a straw. Sucking air out of the straw reduces the pressure inside the straw. The greater atmospheric pressure on the liquid pushes it up into the straw to replace the air that has been sucked out.  5.2 Pressure of a Gas 175 SI Units of Pressure Pressure is one of the most readily measurable properties of a gas. In order to under- stand how we measure the pressure of a gas, it is helpful to know how the units of measurement are derived. We begin with velocity and acceleration. Velocity is defined as the change in distance with elapsed time; that is, distance moved velocity 5 elapsed time The SI unit for velocity is m/s, although we also use cm/s. Acceleration is the change in velocity with time, or change in velocity acceleration 5 elapsed time Acceleration is measured in m/s2 (or cm/s2). The second law of motion, formulated by Sir Isaac Newton† in the late seven- teenth century, defines another term, from which the units of pressure are derived, namely, force. According to this law, force 5 mass 3 acceleration In this context, the SI unit of force is the newton (N), where 1 N is roughly equivalent to the force exerted by Earth’s gravity on an apple. 1 N 5 1 kg m/s2 Finally, we define pressure as force applied per unit area: force pressure 5 area The SI unit of pressure is the pascal (Pa),‡ defined as one newton per square meter: 1 Pa 5 1 N/m2 Atmospheric Pressure The atoms and molecules of the gases in the atmosphere, like those of all other matter, are subject to Earth’s gravitational pull. As a consequence, the atmosphere is much denser near the surface of Earth than at high altitudes. (The air outside the pressurized cabin of an airplane at 9 km is too thin to breathe.) In fact, the density of air decreases very rapidly with increasing distance from Earth. Measurements show that about 50 percent of the atmosphere lies within 6.4 km of Earth’s surface, 90 percent within 16 km, and 99 percent within 32 km. Not surprisingly, the denser the air is, the greater the pressure it exerts. The force experienced by any area exposed to Earth’s atmosphere is equal to the weight of the column of air above it. Atmospheric pressure is the pressure exerted by Earth’s atmosphere (Figure 5.2). The actual value of atmospheric pressure depends on location, temperature, and Column of air weather conditions. † Sir Isaac Newton (1642–1726). English mathematician, physicist, and astronomer. Newton is regarded by many as one of the two greatest physicists the world has known (the other is Albert Einstein). There was hardly a branch of physics to which Newton did not make a significant contribution. His book Principia, published in 1687, marks a milestone in the history of science. ‡ Blaise Pascal (1623–1662). French mathematician and physicist. Pascal’s work ranged widely in mathemat- Figure 5.2 A column of air ics and physics, but his specialty was in the area of hydrodynamics (the study of the motion of fluids). He extending from sea level to the also invented a calculating machine. upper atmosphere. 176 Chapter 5 Gases Does atmospheric pressure act only downward, as you might infer from its defi- nition? Imagine what would happen, then, if you were to hold a piece of paper tight (with both hands) above your head. You might expect the paper to bend due to the pressure of air acting on it, but this does not happen. The reason is that air, like water, is a fluid. The pressure exerted on an object in a fluid comes from all directions— downward and upward, as well as from the left and from the right. At the molecular level, air pressure results from collisions between the air molecules and any surface 76 cm with which they come in contact. The magnitude of pressure depends on how often and how strongly the molecules impact the surface. It turns out that there are just as Atmospheric pressure many molecules hitting the paper from the top as there are from underneath, so the paper stays flat. How is atmospheric pressure measured? The barometer is probably the most familiar instrument for measuring atmospheric pressure. A simple barometer con- sists of a long glass tube, closed at one end and filled with mercury. If the tube is carefully inverted in a dish of mercury so that no air enters the tube, some mercury will flow out of the tube into the dish, creating a vacuum at the top Figure 5.3 A barometer for (Figure 5.3). The weight of the mercury remaining in the tube is supported by measuring atmospheric pressure. Above the mercury in the tube is a atmospheric pressure acting on the surface of the mercury in the dish. Standard vacuum. The column of mercury is atmospheric pressure (1 atm) is equal to the pressure that supports a column of supported by the atmospheric mercury exactly 760 mm (or 76 cm) high at 0°C at sea level. In other words, the pressure. standard atmosphere equals a pressure of 760 mmHg, where mmHg represents the pressure exerted by a column of mercury 1 mm high. The mmHg unit is also called the torr, after the Italian scientist Evangelista Torricelli,† who invented the barometer. Thus, 1 torr 5 1 mmHg and 1 atm 5 760 mmHg (exactly) The relation between atmospheres and pascals (see Appendix 2) is 1 atm 5 101,325 Pa 5 1.01325 3 105 Pa and because 1000 Pa 5 1 kPa (kilopascal) 1 atm 5 1.01325 3 102 kPa Examples 5.1 and 5.2 show the conversion from mmHg to atm and kPa. Example 5.1 The pressure outside a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore, the air inside the cabin must be pressurized to protect the passengers. What is the pressure in atmospheres in the cabin if the barometer reading is 688 mmHg? Strategy Because 1 atm 5 760 mmHg, the following conversion factor is needed to obtain the pressure in atmospheres: 1 atm 760 mmHg (Continued) † Evangelista Torricelli (1608–1674). Italian mathematician. Torricelli was supposedly the first person to recognize the existence of atmospheric pressure.  5.2 Pressure of a Gas 177 Solution The pressure in the cabin is given by 1 atm pressure 5 688 mmHg 3 760 mmHg 5 0.905 atm Similar problem: 5.13. Practice Exercise Convert 749 mmHg to atmospheres. Example 5.2 Hurricane Sandy (“Superstorm Sandy”) was one the most destructive hurricanes in recent years, affecting the Caribbean, Cuba, the Bahamas, and 24 states along U.S. east coast. The lowest pressure recorded for Hurricane Sandy was 705 mmHg. What was the pressure in kPa? Strategy Here we are asked to convert mmHg to kPa. Because 1 atm 5 1.01325 3 105 Pa 5 760 mmHg the conversion factor we need is 1.01325 3 105 Pa 760 mmHg Solution The pressure in kPa is 1.01325 3 105 Pa pressure 5 705 mmHg 3 760 mmHg 5 9.40 3 104 Pa 5 94.0 kPa Similar problem: 5.14. Practice Exercise Convert 295 mmHg to kilopascals. Review of Concepts Rank the following pressures from lowest to highest: (a) 736 mmHg, (b) 0.928 atm, (c) 728 torr, (d) 1.12 3 105 Pa. A manometer is a device used to measure the pressure of gases other than the atmosphere. The principle of operation of a manometer is similar to that of a barom- eter. There are two types of manometers, shown in Figure 5.4. The closed-tube manometer is normally used to measure pressures below atmospheric pressure [Fig- ure 5.4(a)], whereas the open-tube manometer is better suited for measuring pressures equal to or greater than atmospheric pressure [Figure 5.4(b)]. Nearly all barometers and many manometers use mercury as the working fluid, despite the fact that it is a toxic substance with a harmful vapor. The reason is that mercury has a very high density (13.6 g/mL) compared with most other liquids. Because the height of the liquid in a column is inversely proportional to the liquid’s density, this property enables the construction of manageably small barometers and manometers. Review of Concepts Would it be easier to drink water with a straw on top or at the foot of Mt. Everest? 178 Chapter 5 Gases Figure 5.4 Two types of manometers used to measure gas pressures. (a) Gas pressure Vacuum may be less or greater than atmospheric pressure. (b) Gas pressure is greater than atmospheric pressure. h h Gas Gas Mercury Pgas = Ph Pgas = Ph + Patm (a) (b) 5.3 The Gas Laws The gas laws we will study in this chapter are the product of countless experiments on the physical properties of gases that were carried out over several centuries. Each of these generalizations regarding the macroscopic behavior of gaseous substances represents a milestone in the history of science. Together they have played a major role in the development of many ideas in chemistry. The Pressure-Volume Relationship: Boyle’s Law Animation In the seventeenth century, Robert Boyle† studied the behavior of gases systemati- The Gas Laws cally and quantitatively. In one series of studies, Boyle investigated the pressure- volume relationship of a gas sample. Typical data collected by Boyle are shown in Table 5.2. Note that as the pressure (P) is increased at constant temperature, the volume (V ) occupied by a given amount of gas decreases. Compare the first data point with a pressure of 724 mmHg and a volume of 1.50 (in arbitrary unit) to the last data point with a pressure of 2250 mmHg and a volume of 0.58. Clearly there is an inverse relationship between pressure and volume of a gas at constant tem- perature. As the pressure is increased, the volume occupied by the gas decreases. Conversely, if the applied pressure is decreased, the volume the gas occupies increases. This relationship is now known as Boyle’s law, which states that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas. † Robert Boyle (1627–1691). British chemist and natural philosopher. Although Boyle is commonly associ- ated with the gas law that bears his name, he made many other significant contributions in chemistry and physics. Despite the fact that Boyle was often at odds with scientists of his generation, his book The Skeptical Chymist (1661) influenced generations of chemists. Table 5.2 Typical Pressure-Volume Relationship Obtained by Boyle P (mmHg) 724 869 951 998 1230 1893 2250 V (arbitrary units) 1.50 1.33 1.22 1.18 0.94 0.61 0.58 PV 1.09 3 103 1.16 3 103 1.16 3 103 1.18 3 103 1.2 3 103 1.2 3 103 1.3 3 103  5.3 The Gas Laws 179 Figure 5.5 Apparatus for studying the relationship between pressure and volume of a gas. (a) The levels of mercury are equal and the pressure of the gas is equal to the atmospheric pressure (760 mmHg). The gas volume is 100 mL. (b) Doubling the pressure by adding more mercury reduces the gas volume 1520 mmHg to 50 mL. (c) Tripling the pressure decreases the gas volume to one-third of the original value. The temperature and amount of gas are kept constant. 760 mmHg Gas 33 mL 50 mL 100 mL Hg (a) (b) (c) The apparatus used by Boyle in this experiment was very simple (Figure 5.5). The pressure applied to a gas is equal to the gas pressure. In Figure 5.5(a), the pressure exerted on the gas is equal to atmospheric pressure and the volume of the gas is 100 mL. (Note that the tube is open at the top and is therefore exposed to atmospheric pressure.) In Figure 5.5(b), more mercury has been added to double the pressure on the gas, and the gas volume decreases to 50 mL. Tripling the pressure on the gas decreases its volume to a third of the original value [Figure 5.5(c)]. We can write a mathematical expression showing the inverse relationship between pressure and volume: 1 Pr V where the symbol r means proportional to. We can change r to an equals sign and write 1 P 5 k1 3 (5.1a) V where k1 is a constant called the proportionality constant. Equation (5.1a) is the mathematical expression of Boyle’s law. We can rearrange Equation (5.1a) and obtain PV 5 k1 (5.1b) This form of Boyle’s law says that the product of the pressure and volume of a gas at constant temperature and amount of gas is a constant. The top diagram in Figure 5.6 is a schematic representation of Boyle’s law. The quantity n is the number of moles of the gas and R is a constant to be defined in Section 5.4. We will see in Section 5.4 that the proportionality constant k1 in Equations (5.1) is equal to nRT. The concept of one quantity being proportional to another and the use of a proportionality constant can be clarified through the following analogy. The daily income of a movie theater depends on both the price of the tickets (in dollars per 180 Chapter 5 Gases Increasing or decreasing the volume of a gas at a constant temperature P P P Volume decreases Volume increases Boyle’s Law (Pressure increases) (Pressure decreases) Boyle’s Law P = (nRT) 1 nRT is constant V Heating or cooling a gas at constant pressure P P P Lower temperature Higher temperature (Volume decreases) (Volume increases) Charles’ Law V = (nR) T nR is constant P P Charles’ Law Heating or cooling a gas at constant volume P P P Lower temperature Higher temperature (Pressure decreases) (Pressure increases) Charles’ Law P = (nR nR V ) T V is constant Dependence of volume on amount of gas at constant temperature and pressure P P P Gas cylinder Remove gas Add gas molecules (Volume decreases) (Volume increases) Valve Avogadro’s Law V = (RT ) n RT is constant P P Figure 5.6 Schematic illustrations of Boyle’s law, Charles’ law, and Avogadro’s law.  5.3 The Gas Laws 181 Figure 5.7 Graphs showing variation of the volume of a gas with the pressure exerted on the gas, at constant temperature. P P (a) P versus V. Note that the volume of the gas doubles as the 0.6 atm pressure is halved. (b) P versus 1yV. The slope of the line is equal to k1. 0.3 atm 2L 4L V 1 –– V (a) (b) ticket) and the number of tickets sold. Assuming that the theater charges one price for all tickets, we write income 5 (dollar/ticket) 3 number of tickets sold Because the number of tickets sold varies from day to day, the income on a given day is said to be proportional to the number of tickets sold: income r number of tickets sold 5 C 3 number of tickets sold where C, the proportionality constant, is the price per ticket. Figure 5.7 shows two conventional ways of expressing Boyle’s findings graphi- cally. Figure 5.7(a) is a graph of the equation PV 5 k1; Figure 5.7(b) is a graph of the equivalent equation P 5 k1 3 1yV. Note that the latter is a linear equation of the form y 5 mx 1 b, where b 5 0 and m 5 k1. Although the individual values of pressure and volume can vary greatly for a given sample of gas, as long as the temperature is held constant and the amount of the gas does not change, P times V is always equal to the same constant. Therefore, for a given sample of gas under two different sets of conditions at constant tem- perature, we have P1V1 5 k1 5 P2V2 or P1V1 5 P2V2 (5.2) where V1 and V2 are the volumes at pressures P1 and P2, respectively. The Temperature-Volume Relationship: Charles’ and Gay-Lussac’s Law Boyle’s law depends on the temperature of the system remaining constant. But sup- pose the temperature changes: How does a change in temperature affect the volume and pressure of a gas? Let us first look at the effect of temperature on the volume of a gas. The earliest investigators of this relationship were French scientists, Jacques Charles† and Joseph Gay-Lussac.‡ Their studies showed that, at constant pressure, the † Jacques Alexandre Cesar Charles (1746–1823). French physicist. He was a gifted lecturer, an inventor of scientific apparatus, and the first person to use hydrogen to inflate balloons. ‡ Joseph Louis Gay-Lussac (1778–1850). French chemist and physicist. Like Charles, Gay-Lussac was a balloon enthusiast. Once he ascended to an altitude of 20,000 ft to collect air samples for analysis. 182 Chapter 5 Gases volume of a gas sample expands when heated and contracts when cooled (Figure 5.8). The quantitative relations involved in changes in gas temperature and volume turn out Capillary to be remarkably consistent. For example, we observe an interesting phenomenon tubing when we study the temperature-volume relationship at various pressures. At any given pressure, the plot of volume versus temperature yields a straight line. By extending Mercury the line to zero volume, we find the intercept on the temperature axis to be 2273.15°C. At any other pressure, we obtain a different straight line for the volume-temperature Gas plot, but we get the same zero-volume temperature intercept at 2273.15°C (Figure 5.9). (In practice, we can measure the volume of a gas over only a limited temperature Low High range, because all gases condense at low temperatures to form liquids.) temperature temperature In 1848 Lord Kelvin† realized the significance of this phenomenon. He identified Figure 5.8 Variation of the 2273.15°C as absolute zero, theoretically the lowest attainable temperature. Then he volume of a gas sample with set up an absolute temperature scale, now called the Kelvin temperature scale, with temperature, at constant absolute zero as the starting point (see Section 1.7). On the Kelvin scale, one kelvin pressure. The pressure exerted on the gas is the sum of the (K) is equal in magnitude to one degree Celsius. The only difference between the atmospheric pressure and the absolute temperature scale and the Celsius scale is that the zero position is shifted. pressure due to the weight of Important points on the two scales match up as follows: the mercury. Kelvin Scale Celsius Scale Under special experimental conditions, Absolute zero 0K 2273.15°C scientists have succeeded in approaching absolute zero to within a small fraction Freezing point of water 273.15 K 0°C of a kelvin. Boiling point of water 373.15 K 100°C The conversion between °C and K is given on p. 16. In most calculations we will use 273 instead of 273.15 as the term relating K and °C. By convention, we use T to denote absolute (kelvin) temperature and t to indicate temperature on the Celsius scale. The dependence of the volume of a gas on temperature is given by VrT V 5 k2T Remember that temperature must be in V kelvins in gas law calculations. or 5 k2 (5.3) T where k2 is the proportionality constant. Equation (5.3) is known as Charles’ and Gay-Lussac’s law, or simply Charles’ law, which states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute † William Thomson, Lord Kelvin (1824–1907). Scottish mathematician and physicist. Kelvin did important work in many branches of physics. Figure 5.9 Variation of the volume of a gas sample with temperature, at constant pressure. Each line 50 P1 represents the variation at a certain pressure. The pressures increase 40 P2 from P1 to P4. All gases ultimately V (mL) condense (become liquids) if they 30 are cooled to sufficiently low P3 temperatures; the solid portions of the lines represent the temperature 20 –273.15°C region above the condensation P4 point. When these lines are 10 extrapolated, or extended (the dashed portions), they all intersect at 0 the point representing zero volume –300 –200 –100 0 100 200 300 400 and a temperature of 2273.15°C. t (°C)  5.3 The Gas Laws 183 temperature of the gas. Charles’ law is also illustrated in Figure 5.6. We see that the proportionality constant k2 in Equation (5.3) is equal to nRyP. Just as we did for pressure-volume relationships at constant temperature, we can compare two sets of volume-temperature conditions for a given sample of gas at constant pressure. From Equation (5.3) we can write V1 V2 5 k2 5 T1 T2 V1 V2 or 5 (5.4) T1 T2 where V1 and V2 are the volumes of the gas at temperatures T1 and T2 (both in kelvins), respectively. Another form of Charles’ law shows that at constant amount of gas and volume, the pressure of a gas is proportional to temperature PrT P 5 k3T P or 5 k3 (5.5) T From Figure 5.6 we see that k3 5 nRyV. Starting with Equation (5.5), we have P1 P2 5 k3 5 T1 T2 P1 P2 or 5 (5.6) T1 T2 where P1 and P2 are the pressures of the gas at temperatures T1 and T2, respectively. Review of Concepts Compare the changes in volume when the temperature of a gas is increased at constant pressure from (a) 200 K to 400 K and (b) 200°C to 400°C. The Volume-Amount Relationship: Avogadro’s Law The work of the Italian scientist Amedeo Avogadro complemented the studies of Avogadro’s name first appeared in Section 3.2. Boyle, Charles, and Gay-Lussac. In 1811 he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas must be proportional to the number of moles of molecules present; that is, Vrn V 5 k4n (5.7) where n represents the number of moles and k4 is the proportionality constant. Equa- tion (5.7) is the mathematical expression of Avogadro’s law, which states that at constant 184 Chapter 5 Gases Figure 5.10 Volume relationship of gases in a chemical reaction. The ratio of the volumes of molecular hydrogen to molecular nitrogen is 3:1, and that of + ammonia (the product) to molecular hydrogen and molecular nitrogen combined (the reactants) is 2:4, or 1:2. 3H2(g) + N2(g) 2NH3(g) 3 molecules + 1 molecule 2 molecules 3 moles + 1 mole 2 moles 3 volumes + 1 volume 2 volumes pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. From Figure 5.6 we see that k4 5 RTyP. According to Avogadro’s law we see that when two gases react with each other, their reacting volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac). For example, consider the synthesis of ammonia from molec- ular hydrogen and molecular nitrogen: 3H2 (g) 1 N2 (g) ¡ 2NH3 (g) 3 mol 1 mol 2 mol Because, at the same temperature and pressure, the volumes of gases are directly proportional to the number of moles of the gases present, we can now write 3H2 (g) 1 N2 (g) ¡ 2NH3 (g) 3 volumes 1 volume 2 volumes The volume ratio of molecular hydrogen to molecular nitrogen is 3:1, and that of ammonia (the product) to the sum of the volumes of molecular hydrogen and molec- ular nitrogen (the reactants) is 2:4 or 1:2 (Figure 5.10). Worked examples illustrating the gas laws are presented in Section 5.4. 5.4 The Ideal Gas Equation Let us summarize the gas laws we have discussed so far: 1 Boyle’s law: V r (at constant n and T ) P Charles’ law: V r T (at constant n and P) Avogadro’s law: V r n (at constant P and T ) We can combine all three expressions to form a single master equation for the behav- ior of gases: nT Vr P nT V5R P or PV 5 nRT (5.8) Keep in mind that the ideal gas equation, unlike the gas laws discussed in Section 5.3, applies to systems that do not undergo changes in pressure, volume, where R, the proportionality constant, is called the gas constant. Equation (5.8), temperature, and amount of a gas. which is called the ideal gas equation, describes the relationship among the four  5.4 The Ideal Gas Equation 185 Figure 5.11 A comparison of the molar volume at STP (which is approximately 22.4 L) with a basketball. variables P, V, T, and n. An ideal gas is a hypothetical gas whose pressure-volume- temperature behavior can be completely accounted for by the ideal gas equation. The molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container. Although there is no such thing in nature as an ideal gas, the ideal gas approximation works rather well for most reasonable temperature and pressure ranges. Thus, we can safely use the ideal gas equation to solve many gas problems. Before we can apply the ideal gas equation to a real system, we must evalu- ate the gas constant R. At 0°C (273.15 K) and 1 atm pressure, many real gases behave like an ideal gas. Experiments show that under these conditions, 1 mole of an ideal gas occupies 22.414 L, which is somewhat greater than the volume of a basketball, as shown in Figure 5.11. The conditions 0°C and 1 atm are called standard temperature and pressure, often abbreviated STP. From Equation (5.8) we can write PV The gas constant can be expressed in R5 different units (see Appendix 2). nT (1 atm)(22.414 L) 5 (1 mol)(273.15 K) L ? atm 5 0.082057 K ? mol 5 0.082057 L ? atmyK ? mol The dots between L and atm and between K and mol remind us that both L and atm are in the numerator and both K and mol are in the denominator. For most calculations, we will round off the value of R to three significant figures (0.0821 L ? atmyK ? mol) and use 22.41 L for the molar volume of a gas at STP. Example 5.3 shows that if we know the quantity, volume, and temperature of a gas, we can calculate its pressure using the ideal gas equation. Unless otherwise stated, we assume that the temperatures given in °C in calculations are exact so that they do not affect the number of significant figures. Example 5.3 Sulfur hexafluoride (SF6) is a colorless and odorless gas. Due to its lack of chemical reactivity, it is used as an insulator in electronic equipment. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C. (Continued) SF6 186 Chapter 5 Gases Strategy The problem gives the amount of the gas and its volume and temperature. Is the gas undergoing a change in any of its properties? What equation should we use to solve for the pressure? What temperature unit should we use? Solution Because no changes in gas properties occur, we can use the ideal gas equation to calculate the pressure. Rearranging Equation (5.8), we write nRT P5 V (1.82 mol) (0.0821 L ? atmyK ? mol) (69.5 1 273) K 5 5.43 L Similar problem: 5.32. 5 9.42 atm Practice Exercise Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at 6.54 atm and 76°C. By using the fact that the molar volume of a gas occupies 22.41 L at STP, we can calculate the volume of a gas at STP without using the ideal gas equation. Example 5.4 Ammonia gas is used as a refrigerant in food processing and storage industries. Calculate the volume (in liters) occupied by 7.40 g of NH3 at STP. NH3 Strategy What is the volume of one mole of an ideal gas at STP? How many moles are there in 7.40 g of NH3? Solution Recognizing that 1 mole of an ideal gas occupies 22.41 L at STP and using the molar mass of NH3 (17.03 g), we write the sequence of conversions as grams of NH3 ¡ moles of NH3 ¡ liters of NH3 at STP so the volume of NH3 is given by 1 mol NH3 22.41 L V 5 7.40 g NH3 3 3 17.03 g NH3 1 mol NH3 5 9.74 L It is often true in chemistry, particularly in gas-law calculations, that a problem can be solved in more than one way. Here the problem can also be solved by first converting 7.40 g of NH3 to number of moles of NH3, and then applying the ideal gas equation (V 5 nRTyP). Try it. Industrial ammonia refrigeration system. Check Because 7.40 g of NH3 is smaller than its molar mass, its volume at STP Similar problem: 5.40. should be smaller than 22.41 L. Therefore, the answer is reasonable. Practice Exercise What is the volume (in liters) occupied by 49.8 g of HCl at STP? Review of Concepts Assuming ideal behavior, which of the following gases will have the greatest volume at STP? (a) 0.82 mole of He. (b) 24 g of N2. (c) 5.0 3 1023 molecules of Cl2. Which gas will have the greatest density?  5.4 The Ideal Gas Equation 187 The ideal gas equation is useful for problems that do not involve changes in P, V, T, and n for a gas sample. Thus, if we know any three of the variables we can calculate the fourth one using the equation. At times, however, we need to deal with changes in pressure, volume, and temperature, or even in the amount of gas. When conditions change, we must employ a modified form of the ideal gas equation that takes into account the initial and final conditions. We derive the modified equation as follows. From Equation (5.8), P1V1 P2V2 The subscripts 1 and 2 denote the initial R5 (before change) and R5 (after change) and final states of the gas, respectively. n1T1 n2T2 Therefore, P1V1 P2V2 5 (5.9) n1T1 n2T2 It is interesting to note that all the gas laws discussed in Section 5.3 can be derived from Equation (5.9). If n1 5 n2, as is usually the case because the amount of gas normally does not change, the equation then becomes P1V1 P2V2 5 (5.10) T1 T2 Applications of Equation (5.9) are shown in Examples 5.5, 5.6, and 5.7. Example 5.5 An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? Strategy The amount of gas inside the balloon and its temperature remain constant, but both the pressure and the volume change. What gas law do you need? Solution We start with Equation (5.9) P1V1 P2V2 5 n1T1 n2T2 Because n1 5 n2 and T1 5 T2, P1V1 5 P2V2 A scientific research helium which is Boyle’s law [see Equation (5.2)]. The given information is tabulated: balloon. Initial Conditions Final Conditions P1 5 1.0 atm P2 5 0.40 atm V1 5 0.55 L V2 5 ? Therefore, P1 V 2 5 V1 3 P2 1.0 atm 5 0.55 L 3 0.40 atm 5 1.4 L (Continued) 188 Chapter 5 Gases Check When pressure applied on the balloon is reduced (at constant temperature), the helium gas expands and the balloon’s volume increases. The final volume is greater Similar problem: 5.19. than the initial volume, so the answer is reasonable. Practice Exercise A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. Calculate the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL. Example 5.6 Argon is an inert gas used in lightbulbs to retard the vaporization of the tungsten filament. A certain lightbulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure (in atm). Strategy The temperature and pressure of argon change but the amount and volume of gas remain the same. What equation would you use to solve for the final pressure? What temperature unit should you use? Solution Because n1 5 n2 and V1 5 V2, Equation (5.9) becomes P1 P2 Electric lightbulbs are usually filled 5 with argon. T1 T2 which is Charles’ law [see Equation (5.6)]. Next we write Initial Conditions Final Conditions Remember to convert °C to K when P1 5 1.20 atm P2 5 ? solving gas-law problems. T1 5 (18 1 273) K 5 291 K T2 5 (85 1 273) K 5 358 K The final pressure is given by T2 One practical consequence of this P2 5 P1 3 relationship is that automobile tire T1 pressures should be checked only when 358 K the tires are at normal temperatures. 5 1.20 atm 3 After a long drive (especially in the 291 K summer), tires become quite hot, and 5 1.48 atm the air pressure inside them rises. Check At constant volume, the pressure of a given amount of gas is directly proportional Similar problem: 5.36. to its absolute temperature. Therefore the increase in pressure is reasonable. Practice Exercise A sample of oxygen gas initially at 0.97 atm is cooled from 21°C to 268°C at constant volume. What is its final pressure (in atm)? Example 5.7 A small bubble rises from the bottom of a lake, where the temperature and pressure are 8°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the pressure is 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL. (Continued)  5.4 The Ideal Gas Equation 189 Strategy In solving this kind of problem, where a lot of information is given, it is sometimes helpful to make a sketch of the situation, as shown here: What temperature unit should be used in the calculation? Solution According to Equation (5.9) P1V1 P2V2 5 n1T1 n2T2 We assume that the amount of air in the bubble remains constant, that is, n1 5 n2 so that P1V1 P2V2 5 T1 T2 which is Equation (5.10). The given information is summarized: Initial Conditions Final Conditions P1 5 6.4 atm P2 5 1.0 atm We can use any appropriate units for V1 5 2.1 mL V2 5 ? volume (or pressure) as long as we use the same units on both sides of the equation. T1 5 (8 1 273) K 5 281 K T2 5 (25 1 273) K 5 298 K Rearranging Equation (5.10) gives P1 T2 V2 5 V1 3 3 P2 T1 6.4 atm 298 K 5 2.1 mL 3 3 1.0 atm 281 K 5 14 mL Check We see that the final volume involves multiplying the initial volume by a ratio of pressures (P1yP2) and a ratio of temperatures (T2yT1). Recall that volume is inversely proportional to pressure, and volume is directly proportional to temperature. Because the pressure decreases and temperature increases as the bubble rises, we expect the bubble’s volume to increase. In fact, here the change in pressure plays a greater role in the volume change. Similar problem: 5.35. Practice Exercise A gas initially at 4.0 L, 1.2 atm, and 66°C undergoes a change so that its final volume and temperature are 1.7 L and 42°C. What is its final pressure? Assume the number of moles remains unchanged. Density Calculations If we rearrange the ideal gas equation, we can calculate the density of a gas: n P 5 V RT 190 Chapter 5 Gases The number of moles of the gas, n, is given by m n5 m where m is the mass of the gas in grams and m is its molar mass. Therefore m P 5 mV RT Because density, d, is mass per unit volume, we can write m Pm d5 5 (5.11) V RT Unlike molecules in condensed matter (that is, in liquids and solids), gaseous mole- cules are separated by distances that are large compared with their size. Consequently, the density of gases is very low under atmospheric conditions. For this reason, gas densities are usually expressed in grams per liter (g/L) rather than grams per milliliter (g/mL), as Example 5.8 shows. Example 5.8 CO2 Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at 0.990 atm and 55°C. Strategy We need Equation (5.11) to calculate gas density. Is sufficient information provided in the problem? What temperature unit should be used? Solution To use Equation (5.11), we convert temperature to kelvins (T 5 273 1 55 5 328 K) and use 44.01 g for the molar mass of CO2: Pm d5 RT (0.990 atm) (44.01 g/mol) 5 5 1.62 g/L (0.0821 L ? atmyK ? mol) (328 K) Alternatively, we can solve for the density by writing mass density 5 volume Being an intensive property, density is Assuming that we have 1 mole of CO2, the mass is 44.01 g. The volume of the gas can independent of the amount of substance. be obtained from the ideal gas equation Therefore, we can use any convenient amount to help us solve the problem. nRT V5 P (1 mol) (0.0821 L ? atmyK ? mol) (328 K) 5 0.990 atm 5 27.2 L Therefore, the density of CO2 is given by 44.01 g d5 5 1.62 g/L 27.2 L (Continued)  5.4 The Ideal Gas Equation 191 Comment ln units of grams per milliliter, the gas density is 1.62 3 1023 g/mL, which is a very small number. In comparison, the density of water is 1.0 g/mL and that of gold is 19.3 g/cm3. Similar problem: 5.48. Practice Exercise What is the density (in g/L) of uranium hexafluoride (UF6) at 779 mmHg and 62°C? The Molar Mass of a Gaseous Substance From what we have seen so far, you may have the impression that the molar mass of a substance is found by examining its formula and summing the molar masses of its component atoms. However, this procedure works only if the actual formula of the substance is known. In practice, chemists often deal with substances of unknown or only partially defined composition. If the unknown substance is gaseous, its molar mass can nevertheless be found thanks to the ideal gas equation. All that is needed is an experimentally determined density value (or mass and volume data) for the gas at a known temperature and pressure. By rearranging Equation (5.11) we get dRT m5 (5.12) P Figure 5.12 An apparatus for measuring the density of a gas. In a typical experiment, a bulb of known volume is filled with the gaseous sub- A bulb of known volume is filled stance under study. The temperature and pressure of the gas sample are recorded, and with the gas under study at a certain temperature and pressure. the total mass of the bulb plus gas sample is determined (Figure 5.12). The bulb is First the bulb is weighed, and then evacuated (emptied) and weighed again. The difference in mass is the mass of then it is emptied (evacuated) and the gas. The density of the gas is equal to its mass divided by the volume of the bulb. weighed again. The difference in masses gives the mass of the Once we know the density of a gas, we can calculate the molar mass of the substance gas. Knowing the volume of the using Equation (5.12). bulb, we can calculate the density The mass spectrometer has become the dominant instrument for determining of the gas. Under atmospheric conditions, 100 mL of air weigh molar mass, but the determination of molar mass by the density method is still useful, about 0.12 g, an easily measured as illustrated by Example 5.9. quantity. Example 5.9 A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36°C and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula. Strategy Because Equations (5.11) and (5.12) are rearrangements of each other, we can calculate the molar mass of a gas if we know its density, temperature, and pressure. The molecular formula of the compound must be consistent with its molar mass. What temperature unit should we use? Solution From Equation (5.12) dRT m5 P (7.71 g/L) (0.0821 L ? atmyK ? mol) (36 1 273) K Note that we can determine the molar 5 mass of a gaseous compound by this 2.88 atm procedure without knowing its chemical 5 67.9 g/mol formula. (Continued) 192 Chapter 5 Gases Alternatively, we can solve for the molar mass by writing mass of compound molar mass of compound 5 moles of compound From the given density we know there are 7.71 g of the gas in 1 L. The number of moles of the gas in this volume can be obtained from the ideal gas equation PV n5 RT (2.88 atm) (1.00 L) 5 (0.0821 L ? atmyK ? mol) (309 K) 5 0.1135 mol Therefore, the molar mass is given by mass 7.71 g m5 5 5 67.9 g/mol number of moles 0.1135 mol ClO2 We can determine the molecular formula of the compound by trial and error, using only the knowledge of the molar masses of chlorine (35.45 g) and oxygen (16.00 g). We know that a compound containing one Cl atom and one O atom would have a molar mass of 51.45 g, which is too low, while the molar mass of a compound made up of two Cl atoms and one O atom is 86.90 g, which is too high. Thus, the compound must contain one Cl atom and two O atoms and have the formula ClO2, which has a molar Similar problems: 5.43, 5.47. mass of 67.45 g. Practice Exercise The density of a gaseous organic compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass?